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3 years ago

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Which situation represents a proportional relationship between the number of laps run by each student and their time.desmond's t

ime (s) 146 292 584laps 2 4 8maria's time (s) 150 320 580laps 2 4 6?

2 answers:

mezya [45]3 years ago

3 0

Desmond's times are proportional. for every 2 laps he takes 146
146 / 2 = 73
292 / 4 = 73
584 / 8 = 73

Maria's time are not proportional
150 / 2 = 75
320 / 4 = 80
580 / 6 = 96.67

muminat3 years ago

3 0

Answer:

Desmond's time

Step-by-step explanation:

Given,

Desmond's time (s) 146 292 584

laps 2 4 8

Maria's time (s) 150 320 580

laps 2 4 6

By the first table,

$\frac{146}{2}=\frac{292}{2}=\frac{584}{8}=73$

Thus, the situation of Desmond shows the proportional relationship between time and number of laps,

By the second table,

$\frac{150}{2}=30$

$\frac{320}{4}=8$

$\frac{580}{6}=96.67$

$\implies \frac{150}{2}\neq \frac{320}{4}\neq \frac{580}{6}$

Thus, the situation of Maria does not show the proportional relationship between time and laps.

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The Volume of the given solid using polar coordinate is: $\frac{-1}{6} \int\limits^{2\pi}_ {0} [(60) ^{3/2} \; -(64) ^{3/2} ] d\theta$

V= $\frac{-1}{6} \int\limits^{2\pi}_ {0} [(60) ^{3/2} \; -(64) ^{3/2} ] d\theta$

<h3>What is Volume of Solid in polar coordinates?</h3>

To find the volume in polar coordinates bounded above by a surface z=f(r,θ) over a region on the xy-plane, use a double integral in polar coordinates.

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So, the ellipsoid gives

$4{(x^{2}+ y^{2)} + z^{2} =64$

4( $r^{2}$ ) + $z^{2}$ = 64

$z^{2}$ = 64- 4( $r^{2}$ )

z=± $\sqrt{64-4r^{2} }$

So, the volume of the solid is given by:

V= $\int\limits^{2\pi}_ 0 \int\limits^1_0{} \, [\sqrt{64-4r^{2} }- (-\sqrt{64-4r^{2} })] r dr d\theta$

= $2\int\limits^{2\pi}_ 0 \int\limits^1_0 \, r\sqrt{64-4r^{2} } r dr d\theta$

To solve the integral take, $64-4r^{2}$ = t

dt= -8rdr

rdr = $\frac{-1}{8} dt$

So, the integral $\int\ r\sqrt{64-4r^{2} } rdr$ become

= $\int\ \sqrt{t } \frac{-1}{8} dt$

= $\frac{-1}{12} t^{3/2}$

= $\frac{-1}{12} (64-4r^{2}) ^{3/2}$

so on applying the limit, the volume becomes

V= $2\int\limits^{2\pi}_ {0} \int\limits^1_0{} \, \frac{-1}{12} (64-4r^{2}) ^{3/2} d\theta$

= $\frac{-1}{6} \int\limits^{2\pi}_ {0} [(64-4(1)^{2}) ^{3/2} \; -(64-4(2)^{0}) ^{3/2} ] d\theta$

V = $\frac{-1}{6} \int\limits^{2\pi}_ {0} [(60) ^{3/2} \; -(64) ^{3/2} ] d\theta$

Since, further the integral isn't having any term of $\theta$ .

we will end here.

The Volume of the given solid using polar coordinate is: $\frac{-1}{6} \int\limits^{2\pi}_ {0} [(60) ^{3/2} \; -(64) ^{3/2} ] d\theta$

Learn more about Volume in polar coordinate here:

brainly.com/question/25172004

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